Understanding Relational Query Languages: Key Concepts and Operations

1. Relational Algebra Chpt 4 Xin Zhang Raghu Ramakrishnan

2. Relational Query Languages • Query languages: Allow manipulation and retrieval of data from a database. • Relational model supports simple, powerful QLs: • Strong formal foundation based on logic. • Allows for much optimization. • Query Languages != programming languages! • QLs not expected to be “Turing complete”. • QLs not intended to be used for complex calculations. • QLs support easy, efficient access to large data sets. Raghu Ramakrishnan

3. Formal Relational Query Languages Two mathematical Query Languages form the basis for “real” languages (e.g. SQL), and for implementation: • Relational Algebra: More operational, very useful for representing execution plans. • Relational Calculus: Lets users describe what they want, rather than how to compute it. • Understanding Algebra is key to understanding • SQL, query processing! Raghu Ramakrishnan

4. Preliminaries • A query is applied to relation instances, and the result of a query is also a relation instance. • Schemasof input relations for a query are fixed (but query will run regardless of instance!) • The schema for the resultof a given query is also fixed! Determined by definition of query language constructs. • Positional vs. named-field notation: • Positional notation easier for formal definitions, named-field notation more readable. • Both used in SQL Raghu Ramakrishnan

5. R1 Example Instances • “Sailors” and “Reserves” relations for our examples. • We’ll use positional or named field notation, assume that names of fields in query results are `inherited’ from names of fields in query input relations. S1 S2 Raghu Ramakrishnan

6. Relational Algebra • Basic operations: • Selection ( ) Selects a subset of rows from relation. • Projection ( ) Deletes unwanted columns from relation. • Cross-product( ) Allows us to combine two relations. • Set-difference ( ) Tuples in reln. 1, but not in reln. 2. • Union( ) Tuples in reln. 1 and in reln. 2. • Additional operations: • Intersection, join, division, renaming: Not essential, but (very!) useful. Raghu Ramakrishnan

7. Projection • Deletes attributes that are not in projection list. • Schema of result contains exactly the fields in the projection list, with the same names that they had in the (only) input relation. • Projection operator has to eliminate duplicates! • Note: real systems typically don’t do duplicate elimination unless the user explicitly asks for it. Raghu Ramakrishnan

8. Selection • Selects rows that satisfy selection condition. • No duplicates in result! • Schema of result identical to schema of (only) input relation. • Result relation can be the input for another relational algebra operation! (Operatorcomposition.) Raghu Ramakrishnan

9. Union, Intersection, Set-Difference AUB A-B AnB B-A A B Raghu Ramakrishnan

10. Union, Intersection, Set-Difference • All of these operations take two input relations, which must be union-compatible: • Same number of fields. • `Corresponding’ fields have the same type. • What is the schema of result? Occur in either S1 or S2 or both Occur in S1 but not in S2 Occur in both S1 and S2 Raghu Ramakrishnan

11. È S 3 S 4 Ç - S 3 S 4 S 3 S 4 More Example Instances • Work out on board sid sname rating age S3 23 sandy 7 21.0 32 windy 8 19.5 59 rope 10 18.0 sid sname rating age S4 23 sandy 7 21.0 32 windy 8 19.5 45 jib 5 25.0 68 topsail 10 25.0 Raghu Ramakrishnan

12. È S 3 S 4 Ç - S 3 S 4 S 3 S 4 Union, Intersection, Set-Difference(continued) sid sname rating age 23 sandy 7 21.0 • Answers 32 windy 8 19.5 59 rope 10 18.0 45 jib 5 25.0 68 topsail 10 25.0 sid sname rating age sid sname rating age 23 sandy 7 21.5 59 rope 10 18.0 32 windy 8 19.5 Raghu Ramakrishnan

13. Cross-Product • Each row of S1 is paired with each row of R1. • Result schema has one field per field of S1 and R1, with field names `inherited’ if possible. • Conflict: Both S1 and R1 have a field called sid. • Renaming operator: Raghu Ramakrishnan

14. Joins • Condition Join: • Result schema same as that of cross-product. • Fewer tuples than cross-product, might be able to compute more efficiently • Sometimes called a theta-join. Raghu Ramakrishnan

15. Joins • Equi-Join: A special case of condition join where the condition c contains only equalities. • Result schema similar to cross-product, but only one copy of fields for which equality is specified. • Natural Join: Equijoin on all common fields. Raghu Ramakrishnan

16. Division • Not supported as a primitive operator, but useful for expressing queries like: Find sailors who have reserved allboats. • Let A have 2 fields, x and y; B have only field y: • A/B contains all x tuples (sailors) such that for everyy tuple (boat) in B, there is an xy tuple in A. • Or: If the set of y values (boats) associated with an x value (sailor) in A contains all y values in B, the x value is in A/B. • In general, x and y can be any lists of fields; y is the list of fields in B, and x y is the list of fields of A. Raghu Ramakrishnan

17. Examples of Division A/B B1 B2 B3 A/B2 A/B3 A/B1 A Raghu Ramakrishnan

18. all disqualified tuples Expressing A/B Using Basic Operators • Division is not essential op; just a useful shorthand. • (Also true of joins, but joins are so common that systems implement joins specially.) • Idea: For A/B, compute all x values that are not `disqualified’ by some y value in B. • x value is disqualified if by attaching y value from B, we obtain an xy tuple that is not in A. • Disqualified x values: A/B: Raghu Ramakrishnan

19. sno pno s2 p4 sno pno sno s3 p4 s1 p2 s1 s1 p4 ) - A ) - A ) - A ( ( ( s2 s2 p2 s3 s2 p4 s4 sno s3 p2 p p p p p p s2 A A A A A A sno sno sno sno sno sno s3 p4 s3 s4 p2 p p X B2 X B2 X B2 X B2 s4 p4 )) = ) sno sno ( ( Expressing A/B Using Basic Operators(continued) • A/B2 sno s1 - ( s4 Raghu Ramakrishnan

20. Expressing A/B Using Basic Operators(continued) • Work A/B3 on board Raghu Ramakrishnan

21. sno s1 ) - A ) - A ) - A ( ( ( s2 s3 s4 p p p p p p A A A A A A sno sno sno sno sno sno p p sno X B3 X B3 X B3 X B3 ) )) = sno sno ( ( s1 Expressing A/B Using Basic Operators(continued) sno pno s2 p4 sno pno s3 p1 • A/B3 s1 p1 s3 p4 s1 p2 s1 p4 s4 p1 s2 p1 s2 p2 s2 p4 sno s3 p1 s2 s3 p2 s3 p4 s3 s4 p1 s4 s4 p2 s4 p4 - ( Raghu Ramakrishnan

22. R1 Example Instances B1 S1 bid bname bcolor red 101 Gippy green 103 Fullsail S2 Raghu Ramakrishnan

23. Solution 2: • Solution 3: Find names of sailors who’ve reserved boat #103 • Solution 1: Raghu Ramakrishnan

24. A more efficient solution: Find names of sailors who’ve reserved a red boat • Information about boat color only available in Boats; so need an extra join: • A query optimizer can find this given the first solution! Raghu Ramakrishnan

25. Find sailors who’ve reserved a red or a green boat • Can identify all red or green boats, then find sailors who’ve reserved one of these boats: • Can also define Tempboats using union! • Select boats color=red and union with color=green • What happens if is replaced by in this query? • Nothing selected because color can’t be red & green Raghu Ramakrishnan

26. Find sailors names who’ve reserved a red and a green boat • Previous approach won’t work! Must identify sailors who’ve reserved red boats, sailors who’ve reserved green boats, then find the intersection (note that sid is a key for Sailors): Raghu Ramakrishnan

27. Find the names of sailors who’ve reserved all boats • Uses division; schemas of the input relations to / must be carefully chosen: • To find sailors who’ve reserved all ‘Interlake’ boats: ..... Raghu Ramakrishnan

28. Summary • The relational model has rigorously defined query languages that are simple and powerful. • Relational algebra is more operational; useful as internal representation for query evaluation plans. • Several ways of expressing a given query; a query optimizer should choose the most efficient version. Raghu Ramakrishnan